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    Re: Position from altitude and azimuth.
    From: Frank Reed
    Date: 2020 May 10, 11:33 -0700

    Dave Walden, you wrote:
    "Ho=54°55.9' and AZo=120°27.9'.  Topocentric, unrefracted, center of body.

    Topocentric, unrefracted, center of body. Aha! I can do that in Stellarium with the atmosphere turned off.

    You asked:
    "Where am I?  (Hint: If you use the method I did, you may learn or be reminded of somethings about the definitions of nautical miles.)"

    I didn't "go math" on this, and instead did it by simple search in Stellarium. I started at my own location and then bounced around until the altitude was correct. Oh but the azimuth is wrong. So adjust until that's right. And now the altitude is wrong again. And so on. It converged quickly enough to keep me entertained. Did you do something like that when you experimented with this scenario? You mentioned something about the definition of nautical miles... I'm curious to know how that came into play.

    There's a hybrid method, by the way. It's easy to do part of this by math, but not so easy (unfamiliar at least) to complete it. The easy part (simple law of sines for a spherical triangle) gives you the longitude. From there one can easily do a "hunt and peck" search along that longitude meridian until the altitude and azimuth reach the required values (which they do simultaneously thanks to the now-constrained longitude).

    By the way, my result differed by some seconds of arc (more than +/-6) from the integral values of lat/lon that some folks have posted. I didn't write them down, but they were at least a close match to the numbers that Andrés posted. Was that intentional?

    Frank Reed

       
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